The probability of majority inversion in a two-stage voting system with three states

Abstract

Two-stage voting is prone to majority inversions, a situation in which the outcome of an election is not backed by a majority of popular votes. We study the probability of majority inversion in a model with two candidates, three states and uniformly distributed fractions of supporters for each candidate. The model encompasses equal or distinct population sizes, with equal, population-based or arbitrary voting weights in the second stage. We prove that, when no state can dictate the outcome of the election by commanding a voting weight in excess of one half, the probability of majority inversion increases with the size disparity among the states.

Appendices

Appendix A: Proof of Theorem 1 (probability of inversion)

If $v_{[1]}\le \frac{1}{2}$, then the probability of majority inversion

Each summand can be expressed as a multiple integral

$$\begin{aligned} P_i=\frac{2}{w_1w_2w_3}\mathop {\int \int \int }\limits _{A_i}dxdydz\quad \text{ for }\quad i=1,2,3, \end{aligned}$$

evaluated over the following convex polytopes

$$\begin{aligned} A_1&=\{(x,y,z) : x\in (0,\tfrac{w_1}{2}), y\in (0,\tfrac{w_2}{2}), z\in (0,\tfrac{w_3}{2}), x>y+z\},\\ A_2&=\{(x,y,z) : x\in (0,\tfrac{w_1}{2}), y\in (0,\tfrac{w_2}{2}), z\in (0,\tfrac{w_3}{2}), y>x+z\},\\ A_3&=\{(x,y,z) : x\in (0,\tfrac{w_1}{2}), y\in (0,\tfrac{w_2}{2}), z\in (0,\tfrac{w_3}{2}), z>x+y\}. \end{aligned}$$

Here, we use the fact that if $x_i\sim U(-\frac{1}{2},\frac{1}{2})$, then also $-x_i\sim U(-\frac{1}{2},\frac{1}{2})$ for $i=1,2,3$.

Assume $w_1\ge w_2\ge w_3$. Eliminating y yields

where the transformed integration regions are given by

$$\begin{aligned} A'_1&=\left\{ (x,z) : x\in (0,\tfrac{w_1}{2}), z\in (0,\tfrac{w_3}{2}), 0<x-z<\tfrac{w_2}{2}\right\} ,\\ A''_1&=\left\{ (x,z) : x\in (0,\tfrac{w_1}{2}), z\in (0,\tfrac{w_3}{2}), x-z\ge \tfrac{w_2}{2}\right\} ,\\ A'_2&=\left\{ (x,z) : x\in (0,\tfrac{w_1}{2}), z\in (0,\tfrac{w_3}{2}), x+z<\tfrac{w_2}{2}\right\} ,\\ A'_3&=\left\{ (x,z) : x\in (0,\tfrac{w_1}{2}), z\in (0,\tfrac{w_3}{2}), z-x>0\right\} . \end{aligned}$$

The integrals corresponding to $P_2$ and $P_3$ evaluate as

$$\begin{aligned} P_2&=\frac{2}{w_1w_2w_3}\int ^{\frac{w_3}{2}}_0dz\int ^{\frac{w_2}{2}-z}_0\left( \frac{w_2}{2}-x-z\right) dx\\&=\frac{1}{w_1w_2w_3}\int ^{\frac{w_3}{2}}_0\left( \frac{w_2}{2}-z\right) ^2dz= \frac{3w^2_2-3w_2w_3+w^2_3}{24w_1w_2},\\ P_3&=\frac{2}{w_1w_2w_3}\int ^{\frac{w_3}{2}}_0dz\int ^z_0(z-x)dx=\frac{1}{w_1w_2w_3}\int ^{\frac{w_3}{2}}_0z^2dz=\frac{w^2_3}{24w_1w_2}. \end{aligned}$$

The value of $P_1$ depends on whether the largest weight exceeds $\frac{1}{2}$.

If $w_1>\frac{1}{2}$, then

$$\begin{aligned} P_1&=\frac{2}{w_1w_2w_3}\int ^{\frac{w_3}{2}}_0dz\int ^{z+\frac{w_2}{2}}_ z(x-z)dx+\frac{1}{w_1w_3}\int ^{\frac{w_3}{2}}_0dz\int ^{\frac{w_1}{2}}_{z +\frac{w_2}{2}}dx\\&=\frac{w_2}{4w_1w_3}\int ^{\frac{w_3}{2}}_0dz+\frac{1}{w_1w_3} \int ^{\frac{w_3}{2}}_0\left( \frac{w_1-w_2}{2}-z\right) dz=\frac{2w_1-w_2-w_3}{8w_1}. \end{aligned}$$

If $w_1\le \frac{1}{2}$, then

$$\begin{aligned} P_1&=\frac{2}{w_1w_2w_3}\int ^{\frac{w_1-w_2}{2}}_0dz\int ^{z+\frac{w_2}{2}} _z(x-z)dx+\frac{2}{w_1w_2w_3}\int _ {\frac{w_1-w_2}{2}}^{\frac{w_3}{2}}dz\int ^{\frac{w_1}{2}}_z(x-z)dx\\&\quad +\frac{1}{w_1w_3}\int ^{\frac{w_1-w_2}{2}}_0dz\int ^{\frac{w_1}{2}}_{ z+\frac{w_2}{2}}dx=\frac{w_2}{4w_1w_3}\int ^{\frac{w_1-w_2}{2}}_0dz\\&\quad +\frac{1}{w_1w_2w_3}\int _{\frac{w_1-w_2}{2}}^{\frac{w_3}{2}}\left( \frac{w_1}{2}-z\right) ^2dz\\&\quad +\frac{1}{w_1w_3}\int ^{\frac{w_1-w_2}{2}}_0\left( \frac{w_1-w_2}{2}- z\right) dz=\frac{w_1-w_2}{8w_3}+\frac{w^2_2}{24w_1w_3}-\frac{w^2_1}{24w_2w_3}\\&\quad +\frac{w_1-w_3}{8w_2}+\frac{w^2_3}{24w_1w_2}. \end{aligned}$$

Therefore, if $w_1>\frac{1}{2}$, then

$$\begin{aligned} P(w_1,w_2,w_3)= & {} \frac{3w^2_2-3w_2w_3+2w^2_3}{24w_1w_2}+\frac{2w_1-w_2-w_3}{8w_1}\\= & {} \frac{1}{4}-\frac{w_3}{4w_1}+\frac{w^2_3}{12w_1w_2}, \end{aligned}$$

or else

$$\begin{aligned} P(w_1,w_2,w_3)&=\frac{3w^2_2-3w_2w_3+2w^2_3}{24w_1w_2}+\frac{w_1-w_2}{8w_3} +\frac{w^2_2}{24w_1w_3}-\frac{w^2_1}{24w_2w_3}\\&\quad +\frac{w_1-w_3}{8w_2} +\frac{w^2_3}{24w_1w_2}\\&=\frac{w_1-w_2}{8w_3}+\frac{w_1-w_3}{8w_2}+\frac{w_2-w_3}{8w_1} -\frac{w^2_1}{24w_2w_3}\\&\quad +\frac{w^2_2}{24w_1w_3}+\frac{3w^2_3}{24w_1w_2}. \end{aligned}$$

Consider the case $v_{[1]}>\frac{1}{2}$. As before, let $w_1\ge w_2\ge w_3$.

(i) If $v_{[1]}=v_1$, then the probability of majority inversion

Thus, $P'_2=P_2, P'_3=P_3$. For $P'_1$, we have

$$\begin{aligned} P'_1=\frac{2}{w_1w_2w_3}\mathop {\int \int \int }\limits _Adxdydz, \end{aligned}$$

where $A=\{(x,y,z) : x\in (0,\tfrac{w_1}{2}), y\in (0,\tfrac{w_2}{2}), z\in (0,\tfrac{w_3}{2}), x<y+z\}$. Since $P'_1=\frac{1}{4}-P_1$,

$$\begin{aligned} P(w_1,w_2,w_3)&=\frac{w_2}{4w_1}+\frac{w_3^2}{12w_1w_2} \text{ for } w_1>\frac{1}{2},\\ P(w_1,w_2,w_3)&=\frac{1}{4}-\frac{w_1-w_2}{8w_3}-\frac{w_1-w_3}{8w_2}+\frac{w_2-w_3}{8w_1}+\frac{w^2_1}{24w_2w_3}-\frac{w^2_2}{24w_1w_3}\\&\quad +\frac{w^2_3}{24w_1w_2} \text{ for } w_1\le \frac{1}{2}. \end{aligned}$$

(ii) If $v_{[1]}=v_2$, then the probability of majority inversion

Here $P''_1=P_1, P''_2=\frac{1}{4}-P_2, P''_3=P_3$. Therefore,

$$\begin{aligned} P(w_1,w_2,w_3)&=\frac{1}{2}-\frac{w_2}{4w_1} \text{ for } w_1>\frac{1}{2},\\ P(w_1,w_2,w_3)&=\frac{1}{4}+\frac{w_1-w_2}{8w_3}+\frac{w_1-w_3}{8w_2}-\frac{w_2-w_3}{8w_1}-\frac{w^2_1}{24w_2w_3}\\&\quad +\frac{w^2_2}{24w_1w_3}+\frac{w^2_3}{24w_1w_2} \text{ for } w_1\le \frac{1}{2}. \end{aligned}$$

(iii) If $v_{[1]}=v_3$, then the probability of majority inversion

Here $P'''_1=P_1, P'''_2=P_2, P'''_3=\frac{1}{4}-P_3$. Thus,

$$\begin{aligned} P(w_1,w_2,w_3)&=\frac{1}{2}-\frac{w_3}{4w_1} \text{ for } w_1>\frac{1}{2},\\ P(w_1,w_2,w_3)&=\frac{1}{4}+\frac{w_1-w_2}{8w_3}+\frac{w_1-w_3}{8w_2}+\frac{w_2-w_3}{8w_1}-\frac{w^2_1}{24w_2w_3}\\&\quad +\frac{w^2_2}{24w_1w_3}+\frac{w^2_3}{24w_1w_2} \text{ for } w_1\le \frac{1}{2}. \end{aligned}$$

Appendix B: Proof of Theorem 2 (Schur-convexity)

Assume for simplicity that $w_1\ge w_2\ge w_3$. To prove that $P(w_1,w_2,w_3)$ is Schur-convex, we use Lemma 1 to show that

• (a) $P(w_1,w_2+\varepsilon ,w_3-\varepsilon )$ is an increasing function of $\varepsilon$ in $0\le \varepsilon \le \min \{w_1-w_2,w_3\}$;

• (b) $P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)$ is an increasing function of $\varepsilon$ in $0\le \varepsilon \le w_2-w_3$.

Part (a).

If $w_1>w_2+w_3$, then

$$\begin{aligned} P(w_1,w_2+\varepsilon ,w_3-\varepsilon )=\frac{1}{4}-\frac{w_3-\varepsilon }{4w_1} +\frac{(w_3-\varepsilon )^2}{12w_1(w_2+\varepsilon )}, \end{aligned}$$

and the derivative

$$\begin{aligned} \frac{dP(w_1,w_2+\varepsilon ,w_3-\varepsilon )}{d\varepsilon } =\frac{1}{12w_1}\left( 4-\frac{(w_2+w_3)^2}{(w_2+\varepsilon )^2}\right) >0. \end{aligned}$$

The inequality follows since $\frac{w_2+w_3}{w_2+\varepsilon }<2$ due to $w_2-w_3+2\varepsilon >0$.

If $w_1\le w_2+w_3$, then from

$$\begin{aligned} P(w_1,w_2+\varepsilon ,w_3-\varepsilon )&=\frac{w_1-w_2-\varepsilon }{8(w_3-\varepsilon )}+\frac{w_1-w_3+\varepsilon }{8(w_2+\varepsilon )} +\frac{w_2-w_3+2\varepsilon }{8w_1}\\&\quad -\frac{w^2_1}{24(w_2+\varepsilon )(w_3-\varepsilon )}+\frac{(w_2 +\varepsilon )^2}{24w_1(w_3-\varepsilon )}\\&\quad +\frac{3(w_3-\varepsilon )^2}{24w_1(w_2 +\varepsilon )} \end{aligned}$$

we obtain

$$\begin{aligned} 24w_1P(w_1,w_2+\varepsilon ,w_3-\varepsilon )&=6w_1-2w_2-10w_3+8\varepsilon \\&\quad +\frac{3w_1(w_1-w_2-w_3)+(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{w_3-\varepsilon }\\&\quad +\frac{3w_1(w_1-w_2-w_3)+3(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{w_2+\varepsilon }, \end{aligned}$$

and the derivative

$$\begin{aligned} 24w_1\frac{dP(w_1,w_2+\varepsilon ,w_3-\varepsilon )}{d\varepsilon }&=8 +\frac{3w_1(w_1-w_2-w_3)+(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{(w_3-\varepsilon )^2}\\&\quad -\frac{3w_1(w_1-w_2-w_3)+3(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{(w_2+\varepsilon )^2}. \end{aligned}$$

The expression in the numerator of the first fraction is non-negative:

$$\begin{aligned} 3w_1(2w_1-1)+(1-w_1)^2-\frac{w^3_1}{1-w_1}=\frac{(1-2w_1)^3}{1-w_1}\ge 0. \end{aligned}$$

Therefore,

$$\begin{aligned}&24w_1\frac{dP(w_1,w_2+\varepsilon ,w_3-\varepsilon )}{d\varepsilon }\\&\quad \ge 8+\frac{3w_1(w_1-w_2-w_3)+(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{(w_2+\varepsilon )^2}\\&\quad -\frac{3w_1(w_1-w_2-w_3)+3(w_2+w_3)^2-\frac{w^3_1}{w_2+w_3}}{(w_2+\varepsilon )^2}=8-\frac{2(w_2+w_3)^2}{(w_2+\varepsilon )^2}>0. \end{aligned}$$

Part (b).

If $w_1+2\varepsilon >w_2+w_3$, then

$$\begin{aligned} P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)=\frac{1}{4}-\frac{w_3}{4(w_1+\varepsilon )} +\frac{w^2_3}{12(w_1+\varepsilon )(w_2-\varepsilon )}, \end{aligned}$$

and

$$\begin{aligned} \frac{dP(w_1+\varepsilon ,w_2-\varepsilon ,w_3)}{d\varepsilon }=\frac{w_3}{4(w_1 +\varepsilon )^2}+\frac{w^2_3(w_1-w_2+2\varepsilon )}{12(w_1+\varepsilon )^2(w_2-\varepsilon )^2}\ge 0. \end{aligned}$$

If $w_1+2\varepsilon \le w_2+w_3$, then

$$\begin{aligned} P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)&=\frac{w_1-w_2+2\varepsilon }{8w_3}+\frac{w_1-w_3+\varepsilon }{8(w_2-\varepsilon )}+\frac{w_2-w_3-\varepsilon }{8(w_1+\varepsilon )}\\&\quad -\frac{(w_1+\varepsilon )^2}{24(w_2-\varepsilon )w_3}+\frac{(w_2-\varepsilon )^2}{24(w_1+\varepsilon )w_3}\\&\quad +\frac{3w^2_3}{24(w_1+\varepsilon )(w_2-\varepsilon )}. \end{aligned}$$

Consequently,

$$\begin{aligned} 24w_3P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)&=4w_1-4w_2-6w_3+8\varepsilon \\&\quad +\frac{3w_3(w_1+w_2-w_3)+\frac{3w^3_3}{w_1+w_2}-(w_1+w_2)^2}{w_2-\varepsilon }\\&\quad +\frac{3w_3(w_1+w_2-w_3)+\frac{3w^3_3}{w_1+w_2}+(w_1+w_2)^2}{w_1+\varepsilon }. \end{aligned}$$

The derivative

$$\begin{aligned}&24w_3\frac{dP(w_1+\varepsilon ,w_2-\varepsilon ,w_3)}{d\varepsilon }\\&\quad =8+\frac{3w_3(w_1+w_2-w_3)+\frac{3w^3_3}{w_1+w_2}-(w_1+w_2)^2}{(w_2-\varepsilon )^2}\\&\qquad -\frac{3w_3(w_1+w_2-w_3)+\frac{3w^3_3}{w_1+w_2}+(w_1+w_2)^2}{(w_1+\varepsilon )^2}\\&\quad =8-\frac{2(w_1+w_2)^2}{(w_2-\varepsilon )^2}+A\left( \frac{1}{(w_2-\varepsilon )^2}-\frac{1}{(w_1+\varepsilon )^2}\right) , \end{aligned}$$

where $A=3w_3(1-2w_3)+\frac{3w^3_3}{1-w_3}+(1-w_3)^2$. We show that

$$\begin{aligned} A\frac{(w_1-w_2+2\varepsilon )(w_1+w_2)}{(w_1+\varepsilon )^2(w_2-\varepsilon )^2}\ge & {} 2\left[ \left( \frac{w_1+\varepsilon }{w_2-\varepsilon }+1\right) ^2-2^2\right] \\= & {} \frac{2(w_1-w_2+2\varepsilon )(w_1+3w_2-2\varepsilon )}{(w_2-\varepsilon )^2}, \end{aligned}$$

or, equivalently,

$$\begin{aligned} 2(w_2-\varepsilon )^2(3(1-w_3)-2(w_2-\varepsilon ))\ge 1-6w_3+12w^2_3-10w^3_3. \end{aligned}$$

The expression on the left-hand side is not smaller than $4(\frac{1}{2}-w_3)^2(1-w_3)$ since $w_2-\varepsilon \ge \frac{1}{2}-w_3$, $2(w_2-\varepsilon )\le 1-w_3$. But $4(\frac{1}{2}-w_3)^2(1-w_3)=1-5w_3+8w^2_3-4w^3_3\ge 1-6w_3+12w^2_3-10w^3_3$, as $(1-5w_3+8w^2_3-4w^3_3)-(1-6w_3+12w^2_3-10w^3_3)=w_3(1-4w_3+6w^2_3)=w_3((1-2w_3)^2+2w^2_3)\ge 0$.

Appendix C: The upper bound on the probability for EC

Again, assume $w_1\ge w_2\ge w_3$. In part (b) of Appendix B we have established that the function $P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)$ is increasing of $\varepsilon$ in $0\le \varepsilon \le w_2-w_3$ for EC if $w_1+2\varepsilon \le w_2+w_3$. For EC, we show that if $w_1+2\varepsilon >w_2+w_3$, the function $P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)$ is decreasing in $\varepsilon$ for $0\le \varepsilon \le w_2-w_3$. Indeed,

$$\begin{aligned} P(w_1+\varepsilon ,w_2-\varepsilon ,w_3)=\frac{w_2-\varepsilon }{4(w_1 +\varepsilon )}+\frac{w^2_3}{12(w_1+\varepsilon )(w_2-\varepsilon )} \end{aligned}$$

implies

$$\begin{aligned} \frac{dP(w_1+\varepsilon ,w_2-\varepsilon , w_3)}{d\varepsilon }=\frac{(w_1-w_2+2\varepsilon )w^2_3-3(w_1+w_2)(w_2-\varepsilon )^2}{12(w_1+\varepsilon )^2(w_2-\varepsilon )^2}. \end{aligned}$$

Denote $X=w_2-\varepsilon$. Let us show that $3(w_1+w_2)X^2+2w^2_3X-(w_1+w_2)w^2_3>0$ for $X\in [w_3,w_2]$. The roots $X_{1,2}$ of the quadratic are

$$\begin{aligned} X_{1,2}=\frac{-w^2_3\pm \sqrt{w^4_3+3(w_1+w_2)^2w^2_3}}{3(w_1+w_2)} \end{aligned}$$

with the largest root being smaller than $w_3$. It follows that $w^*_1=\frac{1}{2}$ maximizes $P(w_1,w_2,w_3)$.

The function

$$\begin{aligned} P\left( \frac{1}{2},w_2,\frac{1}{2}-w_2\right) =\frac{w_2}{2}+\frac{(\frac{1}{2}-w_2)^2}{6w_2},\quad \frac{1}{4}\le w_2\le \frac{1}{2} \end{aligned}$$

is increasing in $w_2$, and so it is maximized for $w^*_2=\frac{1}{2}$, because

$$\begin{aligned} \frac{dP(\frac{1}{2},w_2,\frac{1}{2}-w_2)}{dw_2}=\frac{2}{3}-\frac{1}{24w^2_2}=\frac{2}{3}\left( 1-\frac{1}{16w^2_2}\right)>0 \text{ for } w_2>\frac{1}{4}. \end{aligned}$$

Consequently, $\mathop {{{\mathrm{arg\,max}}}}\limits _{(w_1,w_2,w_3)} P(w_1,w_2,w_3)=(\frac{1}{2},\frac{1}{2},0)$ and $\max \limits _{(w_1,w_2,w_3)} P(w_1,w_2,w_3)=\frac{1}{4}$.